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# 2.1: Axioms and Basic Definitions

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Real numbers can be constructed step by step: first the integers, then the rationals, and finally the irrationals. Here, however, we shall assume the set of all real numbers, denoted $$E^{1},$$ as already given, without attempting to reduce this notion to simpler concepts. We shall also accept without definition (as primitive concepts) the notions of the sum $$(a+b)$$ and the product, $$(a \cdot b)$$ or $$(a b),$$ of two real numbers, as well as the inequality relation $$<$$ (read "less than" ). Note that $$x \in E^{1}$$ means "x is in $$E^{1,},$$ i.e., "x is a real number."

It is an important fact that all arithmetic properties of reals can be deduced from several simple axioms, listed (and named) below.

## Axioms of Addition and Multiplication

Definition

1. (closure laws) The sum $$x+y,$$ and the product $$x y,$$ any real numbers are real numbers themselves. In symbols,

$\left(\forall x, y \in E^{1}\right) \quad(x+y) \in E^{1} \text{ and } (x y) \in E^{1}$

2. (commutative laws)

$\left(\forall x, y \in E^{1}\right) \quad x+y=y+x \text{ and } x y=y x$

3. (associative laws)

$\left(\forall x, y, z \in E^{1}\right) \quad(x+y)+z=x+(y+z) \text{ and } (x y) z=x(y z)$

4. (existence of neutral elements)

(a) There is a (unique) real number, called zero (0), such that, for all real $$x, x+0=x$$.

(b) There is a (unique) real number, called one (1), such that 1 $$1 \neq 0$$ and, for all real $$x, x \cdot 1=x .$$

In symbols,

(a) $\left(\exists ! 0 \in E^{1}\right)\left(\forall x \in E^{1}\right) \quad x+0=x;$

(b) $\left(\exists 1 \in E^{1}\right)\left(\forall x \in E^{1}\right) \quad x \cdot 1=x, 1 \neq 0.$

(The real numbers 0 and 1 are called the neutral elements of addition and multiplication, respectively.)

5. (existence of inverse elements)

(a) For every real $$x,$$ there is a (unique) real, denoted $$-x,$$ such that $$x+(-x)=0$$.

(b) For every real $$x$$ other than $$0,$$ there is a (unique) real, denoted $$x^{-1}$$, such that $$x \cdot x^{-1}=1$$.

In symbols,

(a) $\left(\forall x \in E^{1}\right)\left(\exists !-x \in E^{1}\right) \quad x+(-x)=0;$

(b) $\left(\forall x \in E^{1} | x \neq 0\right)\left(\exists ! x^{-1} \in E^{1}\right) \quad x x^{-1}=1.$

(The real numbers $$-x$$ and $$x^{-1}$$ are called, respectively, the additive inverse (or the symmetric) and the multiplicative inverse (or the reciprocal) of $$x . )$$

6. (distributive law)

$\left(\forall x, y, z \in E^{1}\right) \quad(x+y) z=x z+y z$

## Axioms of Order

Definition

7. (trichotomy) For any real $$x$$ and $$y,$$ we have

$\text{either} x<y \text{ or } y<x \text{ or } x=y$

but never two of these relations together.

8. (transitivity)

$\left(\forall x, y, z \in E^{1}\right) \quad x<y \text{ and } y<z \text{ implies } x<z$

9. (monotonicity of addition and multiplication) For any $$x, y, z \in E^{1}$$, we have

(a) $x<y \text{ implies } x+z<y+z;$

(b) $x<y \text{ and } z>0 \text{ implies ] x z<y z.$

Note 1: The uniqueness assertions in Axioms 4 and 5 are actually redundant since they can be deduced from other axioms. We shall not dwell on this.

Note 2: Zero has no reciprocal; i.e., for no $$x$$ is $$0 x=1 .$$ In fact, $$0 x=0 .$$ For, by Axioms VI and IV,

$0 x+0 x=(0+0) x=0 x=0 x+0.$

Cancelling $$0 x($$ i.e., adding $$-0 x$$ on both sides $$),$$ we obtain $$0 x=0,$$ by Axioms 3 and 5 (a).

Note 3: Due to Axioms 7 and 8, real numbers may be regarded as given in a certain order under which smaller numbers precede the larger ones. (This is why we speak of "axioms of order.") The ordering of real numbers can be visualized by "plotting" them as points on a directed line ("the real axis") in a well-known manner. Therefore, $$E^{1}$$ is also often called "the real axis," and real numbers are called "points"; we say "the point x instead of "the number x.

Observe that the axioms only state certain properties of real numbers without specifying what these numbers are. Thus we may treat the reals as just any mathematical objects satisfying our axioms, but otherwise arbitrary. Indeed, our theory also applies to any other set of objects (numbers or not), provided they satisfy our axioms with respect to a certain relation of order $$(<)$$ and certain operations $$(+)$$ and $$(\cdot),$$ which may, but need not, be ordinary addition and multiplication. Such sets exist indeed. We now give them a name.

Definition 1

A field is any set $$F$$ of objects, with two operations $$(+)$$ and $$( .)$$ defined in it in such a manner that they satisfy Axioms 1-6 listed above (with $$E^{1}$$ replaced by $$F,$$ of course).

If $$F$$ is also endowed with a relation $$<$$ satisfying Axioms 7 to 9, we call $$F$$ an ordered field.

In this connection, postulates 1 to 9 are called axioms of an (ordered) field.
By Definition $$1, E^{1}$$ is an ordered field. Clearly, whatever follows from the axioms must hold not only in $$E^{1}$$ but also in any other ordered field. Thus
we shall henceforth state our definitions and theorems in a more general way, speaking of ordered fields in general instead of $$E^{1}$$ alone.

Definition 2

An element $$x$$ of an ordered field is said to be positive if $$x>0$$ or negative if $$x<0 .$$

Here and below, $$" x>y "$$ means the same as $$" y<x . "$$ We also write $$" x \leq y "$$ for $$" x<y$$ or $$x=y^{\prime \prime} ;$$ similarly for $$" x \geq y. "$$

Definition 3

For any elements $$x, y$$ of a field, we define their difference

$x-y=x+(-y)$

If $$y \neq 0,$$ we also define the quotient of $$x$$ by $$y$$

$\frac{x}{y}=x y^{-1}$

also denoted by $$x / y$$.

Note 4: Division by 0 remains undefined.

Definition 4

For any element $$x$$ of an ordered field, we define its absolute value,

$|x|=\left\{\begin{array}{ll}{x} & {\text { if } x \geq 0 \text { and }} \\ {-x} & {\text { if } x<0}\end{array}\right.$

It follows that $$|x| \geq 0$$ always; for if $$x \geq 0,$$ then

$|x|=x \geq 0$

and if $$x<0,$$ then

$|x|=-x>0 . \quad( \text{ Why? } )$

Moreover,

$-|x| \leq x \leq|x|,$

for,

$\text{if } x \geq 0, \text{ then } |x|=x;$

and

$\text{if } x<0, \text{ then } x<|x| \text{ since } |x|>0.$

Thus, in all cases,

$x \leq|x|.$

Similarly one shows that

$-|x| \leq x.$

As we have noted, all rules of arithmetic (dealing with the four arithmetic operations and inequalities) can be deduced from Axioms 1 through 9 and thus apply to all ordered fields, along with $$E^{1}$$ . We shall not dwell on their deduction, limiting ourselves to a few simple corollaries as examples.

Corollary $$\PageIndex{1}$$

(i) $$a(-b)=(-a) b=-(a b)$$;

$$($$ ii) $$\quad(-a)(-b)=a b$$.

Proof

By Axiom 6,

$a(-b)+a b=a[(-b)+b]=a \cdot 0=0.$

Thus

$a(-b)+a b=0.$

By definition, then, $$a(-b)$$ is the additive inverse of $$a b,$$ i.e.,

$a(-b)=-(a b).$

Similarly, we show that

$$(-a) b=-(a b)$$

and that

$-(-a)=a.$

Finally, (ii) is obtained from (i) when $$a$$ is replaced by $$-a . \square$$

Corollary $$\PageIndex{2}$$

In an ordered field, $$a \neq 0$$ implies

$a^{2}=(a \cdot a)>0$

(Hence $$\hskip 4pt$$ $$1 =1^{2}>0 . )$$

Proof

If $$a>0,$$ we may multiply by $$a($$ Axiom 9(b) to obtain

$a \cdot a>0 \cdot a=0, \text{ i.e., } a^{2}>0.$

If $$a<0,$$ then $$-a>0 ;$$ so we may multiply the inequality $$a<0$$ by $$-a$$ and obtain

$a(-a)<0(-a)=0;$

i.e., by Corollary 1,

$-a^{2}<0,$

whence

$a^{2}>0 \hskip 4pt \square$